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A 70 gram ball rolls off the edge of a table and falls freely to the floor 122 cm below. While in free fall it moves 40 cm in the horizontal direction. At the instant it leaves the edge it is moving only in the horizontal direction. In the vertical direction, at this instant it is moving neither up nor down so its vertical velocity is zero. For the interval of free fall:
• What are its final velocity in the vertical direction and its average velocity in the horizontal direction?
answer/question/discussion: The final velocity in the vertical direction should be equal to vf ^2 = v0^2 + 2a’ds so vf^2 = 0 cm/s^2 + 2 (-9.8 cm/s^2) (122 cm) = 2391.2 cm/s (we know v0 = zero because the problem tells us that the instant the ball leaves the table the vertical velocity is 0 cm/s).
Acceleration of gravity isn't 9.8 cm/s^2. Should be very easy to correct that.
You then get vf^2, not vf, and the units of the calculation you give would not be cm/s. After taking the square root to get vf, you would get cm/s.
Be sure to actually do those unit calculations.
• Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
answer/question/discussion: First we find the magnitude of the vector R for distance to be 122cm^2 + 40 cm ^2 = 16484, for which we find the sqrt = 128.39 cm. We can also find the angle of the vector magnitude by using tan^-1 (122 cm / 40 cm) = 71.84 or 72 degrees.
Good calculations, but you have obtained the straight-line distance and direction, from the initial to the final point. The object doesn't follow a straight-line path.
Since we already know the vertical velocity component to be 2391.2 cm/s, we can find the horizontal velocity by using the equation tan (72) = 2391.2 cm/s / x and solve for x = .00128 cm/s, which sounds about correct since the horizontal velocity is supposed to remain constant at zero.
Horizontal acceleration remains 0. Horizontal velocity is constant, unchanging.
You need to calculate the time of fall, then use this result to calculate the unchanging horizontal velocity.
Then use the (corrected) vertical velocity, and the horizontal velocity, to find the magnitude and angle of the velocity vector.
• What are its speed and direction of motion at this instant?
answer/question/discussion: The speed is then 2391.2 cm/s and the direction is 72 degrees.
• What is its kinetic energy at this instant?
answer/question/discussion: KE = ½ * m * v^2 = ½ * 70 g * (2391.2 cm/s ^2) = 200124310.4 kgcm/s, which when converted to m instead of centimeters and kilograms is 2001.243 joules. It will be – 2001.243 joules because it is on the ground. (Has already gone through the up and down conversion). But, we must take out a PE of 0.83692 joules to get - 2000.4 joules. (shown calculations later).
• What was its kinetic energy as it left the tabletop?
answer/question/discussion: It will be 2001.243 joules bcause the law of conservation of energy dictates that KE is equal and opposite. But, we must take out a PE of 0.83692 joules to get - 2000.4 joules (shown calculations later).
• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: The change in gravitational potential energy would be m * g * y or .070 kg * 9.8 m/s^2 * 1.22 m = 0.83692 joules.
• How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: The KE and PE are related by Wnc = ‘dKE + ‘dPE. Thus, they are directly related to one another.
• How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: The final KE is almost completely in the vertical direction, as we showed by our calculations of velocity.
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This took me about a half an hour.
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You did some good things, but they don't all apply to the question at hand, and there were a couple of errors in detail.
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